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Proof step 3 from proof empty (previous step)

# st0 = 0 → st0 = ∅, # st0[a, d] = 0 ⊦ st0[a, d] = ∅

Heuristic cut applied rule cut formula

And closed the goal!

Rule argument

fma: a ∈ st0

Additional information

Used simplifier rules: size-in: ⊦ a ∈ st → # st[a, d] = # st;%Size-empty: ⊦ # ∅ = 0;%In-empty: ⊦ ¬ a ∈ ∅;%Size-insert: ⊦ ¬ a ∈ st → # st[a, d] = # st + 1;%z-01: ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used formulas: a ∈ st0